Integrand size = 19, antiderivative size = 52 \[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=-\frac {(a c-b c x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a-b x}{2 a}\right )}{4 a^2 b c (1+n)} \]
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Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {70} \[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=-\frac {(a c-b c x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {a-b x}{2 a}\right )}{4 a^2 b c (n+1)} \]
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Rule 70
Rubi steps \begin{align*} \text {integral}& = -\frac {(a c-b c x)^{1+n} \, _2F_1\left (2,1+n;2+n;\frac {a-b x}{2 a}\right )}{4 a^2 b c (1+n)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=-\frac {(a-b x) (c (a-b x))^n \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {a-b x}{2 a}\right )}{4 a^2 b (1+n)} \]
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\[\int \frac {\left (-b c x +a c \right )^{n}}{\left (b x +a \right )^{2}}d x\]
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\[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=\int { \frac {{\left (-b c x + a c\right )}^{n}}{{\left (b x + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=\int \frac {\left (- c \left (- a + b x\right )\right )^{n}}{\left (a + b x\right )^{2}}\, dx \]
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\[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=\int { \frac {{\left (-b c x + a c\right )}^{n}}{{\left (b x + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=\int { \frac {{\left (-b c x + a c\right )}^{n}}{{\left (b x + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a c-b c x)^n}{(a+b x)^2} \, dx=\int \frac {{\left (a\,c-b\,c\,x\right )}^n}{{\left (a+b\,x\right )}^2} \,d x \]
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